Nucleation of Water Droplets Questions and Answers

Answer: A

Molecules must assemble into a cluster large enough to be stable purely by chance. The supersaturation this requires is far beyond anything the atmosphere achieves, so the process is essentially never observed in clouds.

Answer: B

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the figure that settles it. Forming a droplet from vapour alone requires enough molecules to meet by chance and survive the surface energy penalty, which needs supersaturations of several hundred per cent. Natural clouds reach a fraction of one per cent. Condensation in the atmosphere is therefore always heterogeneous, on a pre-existing particle.

Answer: C

A wettable particle allows the embryonic droplet to start at a size already past the critical radius. The energy that would have gone into creating curved surface is largely saved, and the process proceeds at very modest supersaturation.

Answer: D

Correct answer: further growth lowers the free energy rather than raising it

The free energy of a cluster is a competition between a volume term that falls as it grows and a surface term that rises, and their sum has a maximum. The radius at that maximum is the critical radius: below it growth costs energy and above it growth releases energy, so a cluster that reaches it grows spontaneously thereafter.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason is the mechanism. Below the critical radius the surface term dominates - the cluster is nearly all surface - so adding a molecule raises the free energy and the cluster is more likely to shed molecules than gain them. That is precisely why it takes an enormous supersaturation to push clusters past the barrier without help from a nucleus.

Answer: B

Molecules at a strongly curved surface have fewer neighbours holding them in place and escape more readily. The effect grows rapidly as the radius falls, which is precisely what makes very small droplets so hard to sustain.

Answer: C

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason names the Kelvin effect. A molecule at a strongly curved surface has fewer neighbours holding it, so it escapes more readily and the equilibrium vapour pressure over a small droplet exceeds that over a flat surface. The consequence is that in a population of droplets the large grow at the expense of the small, which is the beginning of the size distribution a cloud develops.

Answer: D

The correction to the equilibrium vapour pressure falls away rapidly with increasing radius. Beyond roughly a micrometre a droplet behaves for practical purposes like a flat water surface. Curvature therefore matters only at the very beginning of a droplet's life.

Answer: A

Fewer water molecules are present at the surface to escape, so the equilibrium vapour pressure falls in proportion to the mole fraction of water. The effect opposes curvature and is strongest when the droplet is small.

Answer: B

Both are true and the reason gives the mechanism, and this is why haze droplets exist in subsaturated air. They grow and shrink with the humidity without ever activating into cloud droplets. Atmospheric haze on a humid day is exactly this population made visible.